158_Physics ProblemsTechnical Physics - 160 Circular Motion...

160 Circular Motion and Other Applications of Newton’s LawsSOLUTIONS TO PROBLEMSSection 6.1Newton’s Second Law Applied to Uniform Circular MotionP6.1m=300. kg,r=0800.m. The string will break if the tension exceedsthe weight corresponding to 25.0 kg, soTMgmax..===25 0 9 80245afN.When the 3.00 kg mass rotates in a horizontal circle, the tensioncauses the centripetal acceleration,soTmvrv22...ThenvrTmTT20800 24565 3≤==.......maxafafafms22and0653≤≤v.or0808vms.FIG. P6.1P6.2In Fmvr∑=2, both mand rare unknown but remain constant. Therefore, F∑is proportional to v2and increases by a factor of 18 014 02..FHGIKJas vincreases from 14.0 m/s to 18.0 m/s. The total force at the

This is the end of the preview. Sign up
to
access the rest of the document.